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Unformatted text preview: MATH 4377, ADVANCED LINEAR ALGEBRA, SUMMER 2011, Key to HW#8 1. Problem 1 on Page 179. Solutiton : (a) False. (b) False. (c) True, since (0 ,..., 0) is always a solution for a homog system. (d) False. (e) False. (f) False. (g) True, the solution is x = A 1 b . (h) False. Only for homog system, the solution set is a subspace. 2. Problem 2 (b), (d), (f) Page 180. Solutiton : 2(b): By eliminating x 1 , we get x 2 = 2 3 x 3 , and x 1 = x 3 x 2 = 1 3 x 3 . so the solution is ((1 / 3) x 3 , (2 / 3) x 3 ,x 3 ) = x 3 (1 / 3 , 2 / 3 , 1) for x 3 ∈ R . A basis for the solution is can be { (1 , 2 , 3) } . The dimension of the solution space is 1. 2(d): By adding the first and second equation, we get x 1 = 0, we also get x 2 = x 3 . So the solution set is { (0 ,x 2 ,x 2 ) = x 2 (0 , 1 , 1) ,x 2 ∈ R } . A basis for the solution is can be { (0 , 1 , 1) } . The dimension of the solution space is 1. 2(f): The solution set is { (0 , 0) } , there is no basis for it. The dimension of the solution space is zero. 3. Problem 3 (b), (d), (f) Page 180. Solution : 3(b): First we can find a particular solution for the system which is (2 / 3 , 1 / 3 , 0). Hence, using 2(b), the solution is ( x 1 ,x 2 ,x 3 ) = (2 / 3 , 1 / 3 , 0) + t (1 , 2 , 3) where t ∈ R . 3(d): First we can find a particular solution for the system which is (2 , 1 , 0). Hence, using 2(b), the solution is ( x 1 ,x 2 ,x 3 ) = (2 , 1 , 0) + t (0 , 1 , 1) where t ∈ R ....
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 Summer '08
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 Linear Algebra, Algebra

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